Question

If $\omega$ is a complex cube root of unity and $f(n)=2(1+\omega )\left(1+{\omega }^2\right)+3(2+\omega )\left(2+{\omega }^2\right)+\dots (n+1)(n+\omega )\left(n+{\omega }^2\right)$, then the value of $f(19)$ is

• A. $36150$
• B. $36100$
• C. $36101$
• D. $36119$

Answer 1

Answer: (D)

We have,
$\begin{array}{l}f(n)=2(1+\omega )\left(1+{\omega }^2\right)+3(2+\omega )\left(2+{\omega }^2\right)+\dots (n+1)(n+\omega )\left(n+{\omega }^2\right)\\ \Rightarrow f(n)=\sum (n+1)(n+\omega )\left(n+{\omega }^2\right)\\ \Rightarrow f(n)=\sum (n+1)\left(n^2+n{\omega }^2+n\omega +{\omega }^3\right)\\ \Rightarrow f(n)=\sum (n+1)\left(n^2+n\left({\omega }^2+\omega \right)+1\right)\\ \Rightarrow f(n)=\sum (n+1)\left(n^2-n+1\right)\\ \Rightarrow f(n)=\sum \left(n^3+1\right)\\ \Rightarrow f(n)={\left[\frac{n(n+1)}{2}\right]}^2+n\\ \therefore f(19)=[190{]}^2+19=36119\end{array}$